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2 years ago

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A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 90.0 m above the

glacier at a speed of 180 m/s . You may want to review (Pages 86 - 90) . For help with math skills, you may want to review: Mathematical Expressions Involving Squares For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Dock jumping. Part A How far short of the target should it drop the package? Express your answer using three significant figures with the appropriate units.

1 answer:

romanna [79]2 years ago

6 0

Answer:

d = 771.3m

Explanation:

Let's first calculate the time of flight:

$Yf = Yo+Voy*t-g*t^2/2$ where Voy=0. Solving for t:

$t = \sqrt{2*90/9.8} =4.285s$

Now we calculate the horizontal displacement, wihch is the distance from the target to drop the package:

Xf = d = Vox*t

d = 180*4.285 = 771.3m

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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

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$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

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pychu [463]

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